Codeforces #260 div. 2 C. Boredom

C. Boredom

time limit per test

 1 second

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

Sample test(s)

input

2
1 2

output

2

input

3
1 2 3

output

4

input

9
1 2 1 3 2 2 2 2 3

output

10

Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

题意描述：给出一个n个数的序列，a1,a2,.....,an.每次操作从序列中取出一个数，比如ak，将其从序列中删除，则序列中值等于ak-1或ak+1的数都被顺带删除，且本次操作得分ak。问最多得分是多少。

分析：显然把序列最终全删除得分最高（用反证法）。现在假设把序列中的数按大小排序，所有等于a[i]的数的个数保存在数组index[a[i]]中，删除一个数ak，则ak-1和ak+1都被顺带删除，这意味着剩下的ak只能被主动删除，不可能被顺带删除。于是所有同值的数，只有都被主动删除和都被顺带删除两种可能，类似于01背包问题。值的取值范围是1~10^5，假设从左向右dp，对于值i，如果i-1是被主动删除，则 i 不存在，所以dp[i]=dp[i-1]，如果i-1是被顺带删除，则 i 应该被主动删除，所以dp[i]=dp[i-2]+index[i]*i，综上有状态转移方程：dp[i]=max(dp[i-1],dp[i-2]+index[i]*i).

#include <bits/stdc++.h>

using namespace std;

long long kol[1000000], dp[1000000];

int main()


{
    int n;
    cin>>n;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin>>x;
        kol[x]++;
    }
    dp[0] = 0;
    for (int i = 1; i <= 100000; i++)
    {
        dp[i] = dp[max(i-2,0)]+kol[i]*i;
        if (dp[i-1] > dp[i]) dp[i] = dp[i-1];
    }
    cout<<dp[100000]<<endl;
}